Summary

Let T be a homogeneous polynomial in three variables and with rational
integer coëfficients. As a generalisation of Thue’s equation we consider the
ternary form equation T (x, y, z) = 1 in the integral unknowns x, y, z. We
prove some general results when the degree of T is at most three and make
some modest inroads into the case degT > 3. Generalisations to algebraic
number fields are considered at the same time.

1
Ternary Form Equations
F. Beukers

January 8, 2007

1.1 Introduction
Let K be an algebraic number field, which we assume embedded in C, and
let O be its ring of integers. Let S be a finite set of places of K, including
all infinite ones. An element a ∈ K is called an S-integer if |a|v ≤ 1 for all
v 6∈ S. The set of S-integers in K is denoted by OS . The units in OS are
called S-units, i.e. elements a ∈ K such that |a|v = 1 for all v 6∈ S. The set
of S-units is denoted by OS∗ . This group is known to have rank |S| − 1.
A diophantine equation which has been studied extensively this cen-
tury is the so-called Thue-Mahler equation

F (x, y) ∈ OS∗

where F ∈ OS [X, Y ] is a binary form (=homogeneous polynomial in two
variables) with non-zero discriminant. Actually, we must state the problem
more accurate here. Let F be the fractional OS -ideal generated by the
coefficients of F and consider the equation

(F (x, y)) = F · (x, y)d (B)

where (F (x, y)) denotes the (fractional) OS -ideal generated by F (x, y) and
d is the degree of F . We shall refer to such equations as binary form equa-
tions. The unknowns lie in K. Any two solutions (x1 , y1 ), (x2 , y2 ) which are
projectively equivalent as elements of P1 (K) are considered as equivalent.
When counting solutions we count equivalence classes of solutions.
When degF = 1 there exist infinitely many solutions. When degF = 2
we have two cases. The first is when F has zeros in K and |OS∗ | < ∞ or
F has zeros quadratic over K and no place of S splits in the quadratic
extension. Then there are at most finitely many solutions and examples
are xy = ±1 and x2 + y 2 = ±1 in OS = Z. In the second, remaining,

2
case there exist either no solutions at all or infinitely many, examples being
Pell’s equation for infinitely many solutions and 2x2 − 5y 2 = ±1 having
no solutions in Z. As soon as degF ≥ 3 the results are less trivial. It
is a well-known result of Thue-Siegel-Mahler that there are only finitely
many (equivalence classes of) solutions. Among the most striking results on
binary form equations are Baker’s upper bound for the height of solutions
[Ba, ST] and the Evertse-Bombieri-Schmidt [E1, BS] bounds for the number
of solutions, which depend only on K, degF and |S|.
Let us give a geometrical interpretation of binary form equations. Sup-
pose x, y is a solution to (B). Then we have for any v 6∈ S that v(F −1 F (x, y)) =
(x, y)d , in other words, modulo every prime outside of S the projective point
(x : y) stays away from the points defined by F (X, Y ) = 0. More precisely,
let (F ) be the zero divisor of F and extend P1 \(F ) over Spec(OS ) in the ob-
vious way. Then any solution of (B) yields a section of P1 \(F ) → Spec(OS ).
By abuse of language an equivalence class of solutions of (B) will be called
an S-integral point on P1 \ (F ).
More generally, let C be any geometrically irreducible curve defined
over K and let D be a divisor (over K) with simple points. Suppose we
have a model of C \ D over Spec(OS ) for some S. A theorem of Siegel and
Mahler [La, Chapter 8] then states that there are at most finitely many S-
integral points in the following cases, g(C) = 0 and |D| ≥ 3, g(C) ≥ 1 and
|D| ≥ 1. In addition, when g(C) ≥ 2 we know by Faltings’ theorem that
C(K) is finite. Notice that the binary form equation is an example of the
case g(C) = 0, and |D| ≥ 3 corresponds to degF ≥ 3.
We now turn to the subject of this paper. Let T ∈ K[X, Y, Z] be a
ternary form (=homogeneous polynomial in three variables) which may be
reducible over C, but has no multiple components. Let d be the degree of
T and let T be the fractional OS -ideal generated by the coefficients of T .
Consider the ternary form equation
(T (x, y, z)) = T · (x, y, z)d (T)
in the unknowns x, y, z ∈ OS . Any two solutions which are projectively
equivalent as elements of P2 (K) are considered equivalent. Again, when
counting solutions, we count equivalence classes of solutions.
Just as in the binary case we can give a geometrical interpretation.
Let C be the projective curve defined by T (X, Y, Z) = 0. Extend P2 \ C
over Spec(OS ) in the trivial way. Then any solution of (T) yields a section
of P2 \ C → Spec(OS ). Again we call an equivalence class of solutions to
(T) an S-integral point on P2 \ C.

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Contrary to the binary case, literature on ternary form equations is
very scant. Of the few references I could find, the most significant is from
D.H.Lehmer [Le] on the equation x3 + y 3 + z 3 = 1 in x, y, z ∈ Z. We shall
refer to his results in a few moments. A more recent paper is [Si], containing
many speculations and some results on the asymptotics of numbers of inte-
gral points. Finally we mention a paper by Mordell [Mo] which has some
relevance to the topic.
Let us formulate our first question. As an example, when K = Q and
S = {∞}, consider the equation

xd + y d + z d = 1

in x, y, z ∈ Z and where d is an odd positive integer. For any such d the
triples (t, −t, 1) where t ∈ Z form an infinite set of inequivalent solutions.
Notice however that they all lie on the projective straight line x + y = 0.
This is an example of what we will call an exceptional curve. Very often we
will see that (T) has infinitely many solutions but which are concentrated on
a finite number of algebraic curves. A subset of P2 which is not contained
in a finite union of algebraic curves is said to lie Zariski dense in P2 .

Question 1.1.1 When does the solution set of (T) lie Zariski dense in P2 ?

From Lehmer [Le] it follows that the solutions to x3 +y 3 +z 3 = 1 in x, y, z ∈ Z
lie indeed Zariski dense. In fact, a computer search with the constraints
0 < |x| ≤ |y| ≤ |z|, |y| < 100, 000 yielded 78 solutions. See [GLS] for some
more numerical results and some interesting problems, as well as the recent
paper [HR]. Computer searches for some other cubic ternary form equations
were equally rewarding in a small but continuing trickle of solutions. Com-
pared to these, computer searches in the cases deg(T ) ≥ 4 reveal a true
desert. Discarding the (trivial) solutions on exceptional curves one usually
finds a handfull of small ones (if any) and that is it. This watershed be-
tween the cases deg(T ) is 3 and 4 is predicted by Vojta’s Main Conjecture
[V, Conjecture 3.4.3]. From this conjecture we can deduce the following
conjecture.

Conjecture 1.1.2 When the curve defined by T (X, Y, Z) = 0 has normal
crossings and deg(T ) ≥ 4, the solution set to (T) is contained in a finite set
of plane algebraic curves.

It is probably possible to weaken the condition ‘normal crossing’, but lacking
any evidence we dare not make any improved conjecture.

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To get any grasp on Question 1.1.1 we must get an idea of the shape
of the exceptional curves. Let us consider the curves in P2 which contain
infinitely many solutions of (T) . Let X be such an irreducible curve. First
of all we remark that, containing infinitely many K-rational points, X can
be given by an equation over K. Let X̃ be a normalisation of X and let D
be the divisor on X̃ cut out by the curve C. Since there are infinitely many
S-integral points on X̃ \ D we conclude from the Siegel-Mahler theorem that
g(X̃) = 0 and |D| ≤ 2. In other words, X \X ∩C is the image of a morphism
from P1 \ {0} or P1 \ {0, ∞} to P2 \ C. Forgetting for the moment that X
is defined over K we adopt this property as definition

Definition 1.1.3 Let C be an algebraic curve in P2 defined over C and
possibly reducible. An exceptional curve in P2 \ C is the image of a non-
constant morphism from P1 \ V to P2 \ C, where V = {0} or {0, ∞}.

Before answering Question 1.1.1 it seems appropriate to study the following
geometrical question first.

Question 1.1.4 Let C be a plane projective curve over C without multiple
components. Does the union of all exceptional curves in P2 \ C lie Zariski
dense in P2 ?

Notice that, when deg(C) ≤ 3, the answer is yes. If deg(C) ≤ 2 every
straight line in P2 is exceptional. If deg(C) = 3 and C is non-singular take
the set of tangents at the points of C. If deg(C) = 3 and C is singular take
the pencil of straight lines passing through a singularity. Whether or not a
positive answer to Question 1.1.4 implies a positive answer to Question 1.1.1
is a subtle problem. When deg(C) ≤ 3 we shall deal with this in section
3 and show that under very mild conditions the solutions of (T) lie Zariski
dense.
In section 4 we shall say as much as we can about the cases deg(C) ≥ 4.
The latter cases are notoriously difficult to handle. As an example consider
the equation x5 − y 2 z 3 = 1. Presumably this simple looking equation has
only the trivial solutions with x = 1, but I could not prove it. As a side
remark we note that a number of the form y 2 z 3 is a socalled powerful number,
all of whose primes occur with exponent at least 2. So the equation x5 −
y 2 z 3 = 1 can be reformulated as the question for which x ∈ Z the number
x5 − 1 is powerful. Up to this day not a single polynomial P ∈ Z[X] with
at least three distinct roots is known for which the finiteness of the set of

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powerful values is proved. That P requires at least three distinct roots is
illustrated by x2 − 1 = 8y 2 which is a Pellian equation having infinitely
many solutions. Finally we remark that the finiteness of the set of powerful
values of x(x2 −1) implies the existence of infinitely many primes p for which
2p 6≡ 2 (mod p2 ) (see [Ri, p 270ff]). There are other cases of ternary form
equations which give rise to diophantine problem that look amusing, but
seem extremely hard to crack. Consider for example
(x2 + y 2 + z 2 )yz ∈ OS∗
in x, y, z ∈ OS∗ . This implies that y, z ∈ OS∗ and x2 + y 2 + z 2 ∈ OS∗ . Hence
x2 is the sum of three S-units. The difficult question seems to be, how often
can the sum of three S-units be a square? One more example, consider
x5 + y 5 − zx2 y 2 = 1 x, y, z ∈ Z.
This is equivalent to the problem, find x, y ∈ Z such that both x5 ≡ 1(mod
y 2 ) and y 5 ≡ 1(modx2 ).
Finally, in section 5 we collect some results on the set of exceptional
curves corresponding to a given curve C.
Remark. There are a number of topics in diophantine equations which
are closely related to ternary form equations. We mention for example the
work of Manin et al where they count solutions of height bounded by H of
x3 + y 3 + z 3 = 1 and similar equations. Asymptotic results of these counting
procedures can be found in [MaTsch]. It might be of interest to know what
the asymptotics for integral solutions of x3 + y 3 + z 3 = 1 might be. There
is also the remark that when studying T (x, y, z) = 1 we are actually looking
at points on the projective surface T (x, y, z) = wd which are integral with
respect to w = 0. When d = 4 and the surface is a K3-surface, there can
be a Zariski-dense set of rational points on it, which conjecturally does not
happen for integral points. Finally one can apply the ABC-conjecture to
three term equations like x5 − y 2 z 3 = 1 and see what can be expected.
Acknowledgement. I would like to thank Frans Oort for goading me into
a geometrical approach of the subject, which turned out to be quite illumi-
nating.

1.2 Technical preparations
In this section we gather some results which are a basic tool in the sections
that follow. The symbols K, S, OS , OS∗ have the same meaning as in

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the introduction. The notation (α, β, γ) stands for the fractional OS -ideal
generated by α, β, γ ∈ K. To avoid confusion with coordinates of a point in
P2 we denote a projective point by (α : β : γ).

Remark. This theorem does not immediately generalise to rational curves
of higher degree. Consider for example the conic x2 − 2y(y + 5z) = 0 and
the point P = (0 : 0 : 1). We take K = Q and S = {∞}. Note that any
rational point reduces to P modulo 5. Hence there cannot be any S-integral
points on the conic minus P .

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If however (j1 + a)/q 6∈ J we adapt our choice of j1 , j2 as follows.
Let γ ∈ J ∩ pq J. Then j1 − γ and j2 + pq γ are both in J and they satisfy
p(j1 −γ)+q(j2 + pq γ) = c. It thus remains to show that there exists γ ∈ J ∩ pq J
such that (j1 − γ + a)/q ∈ J. In other words, we must show that

j1 + a
∈ J + p−1 J ∩ q −1 J. (2)
q

Notice that the fractional ideal J + p−1 J ∩ q −1 J is the set of α ∈ K such
that
v(α) ≥ min(v(J), v(J) + max(−v(p), −v(q)))
for every valuation v 6∈ S. We must show that v((j1 + a)/q) satisfies these
inequalities. Suppose v(p) ≥ v(q). Then

Proof. This is quite straightforward and we sketch it only in case iii). Let
¯ be such that λ̄ is the conjugate of λ which lies in the field generated
λ̄ and λ̄
by the coordinates of B, and similarly for λ̄¯ . Let M be the matrix spanned
by the coordinates of A, B, C written columnwise. Let L be the diagonal
¯ ). We simply take
matrix diag(λ, λ̄, λ̄
T = MLM−1 .
First, T is independent of the choice of coordinates of A, B, C. Secondly,
because of Galois invariance and because we took λ ≡ 1 (mod δ) the co-
¯ ∈ O∗ .
ordinates of T are in OS . Thirdly, det(T ) = λλ̄λ̄ S

Theorem 1.2.3 Let C be a line or an irreducible conic in P2 , defined over
K. Let A, B ∈ C be distinct points. We distinguish the following cases,
i) A, B are defined over K.
ii) A is defined over a quadratic extension M of K and B is its conjugate.
Let V be the set of S-integral points on C \ {A, B}. Then V is finite in the
following cases: case i) and OS∗ finite, case ii) and no place of S splits in
M . In all other cases V is either empty or infinite.

Remark. To mention a very simple case, take K = Q, S = ∞ and take for
C the straight line Z = 0. For A and B take (0 : 1 : 0) and (1 : 0 : 0). A
point (x : y : 0), x, y ∈ Z can only be S-integral on C \ {A, B} if xy = ±1,
i.e. x and y contain no prime divisors. So the only possibilities are (1 : 1 : 0)
and (1 : −1 : 0). If, on the other hand, we take S = {∞, p} for some prime
p then any point of the form (1 : ±pk : 0), k ∈ Z is S-integral. So there
exist infinitely many.

9
√
Remark. Let M = K( d). To get an idea of the splitting condition for
elements of S we remark that there exists an infinite valuation of K which
splits in M/K unless K is totally real and d is totally negative. To mention
a very simple case take K√= Q, S = ∞ √ and take for C the straight line
Z = 0. For A and B take ( 2 : 1 : 0) and ( 2 : −1 : 0). Any point (x : y : 0)
with x, y ∈ Z satisfying x2 − 2y 2 = 1 can be seen to be an S-integral point
on L \ {A, B}.√So there are infinitely√many of them. If, on the other hand,
we take A = ( −1 : 1 : 0) and B = ( −1 : −1 : 0) any S-integral point on
L \ {A, B} must have the form (x : y : 0) with x, y ∈ Z and x2 + y 2 = 1. So
there exist only finitely many.

Proof. Let us first construct an auxiliary point P . In case C is a straight
line we take for P any K-rational point outside of C. In case C is a conic we
take for P the point of intersection of the tangents at C through the points
A and B. Note that P is K-rational in both cases. Consider the function

u(x) = det(A, P, x)/ det(B, P, x))

on P2 . When C is a line, u(x) is a coordinate on C, when C is a conic, u(x)
has order 2 on C.
Suppose in case i) that OS∗ is infinite and in case ii) that at least one
place of S splits in M . Consider the pencil of conics

α det(A, P, x) det(B, P, x) + β det(A, B, x)2 = 0.

In particular, when C is a conic, it is a member of this pencil. We now
choose units λ, µ, ν as follows. In case i) we take them to be S-units, in
case ii) we take λ an S-unit, µ an S 0 -unit and ν its conjugate. Moreover,
they should be in the subgroups U, U 0 as prescribed by Proposition 1.2.2.
Finally we want that λ2 = µν. Construct the projective automorphism T
as in Proposition 1.2.2 so that we have

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Hence, because of λ2 = µν, T stabilises the conics of our pencil, in particular
C.
Notice that u(T x) = (ν/µ)u(x). Suppose that V is not empty and
that x0 ∈ V . In particular, u(T x0 ) = (ν/µ)T (x0 ). The point T x0 is again
S-integral on C \ {A, B}. Because of our assumption that OS∗ infinite in case
i) and at least one v ∈ S splits in M in case ii), there are infinitely many
choices for ν/µ and hence infinitely many S-rational points on C \ {A, B}.
Suppose in case i) that OS∗ is finite and in case ii) that no place of
S splits in M . Let x be an S-integral point on C \ {A, B} and put x =
αA + βB + γP . Denote by A, B, P the ideals generated by the coefficients
of A, B, P respectively. The condition that x does not reduce to A for any
v 6∈ S implies that for every v 6∈ S,
|αA|v ≤ max(|βB|v , |γP|v ).
Similarly
|βB|v ≤ max(|αA|v , |γP|v ).
When C is a straight line we have γ = 0 and |αA|v = |βB|v for every
v 6∈ S. Notice also that |u(x)|v = |β/α|v . Hence, if C is a straight line,
|u(x)|v = |A/B|v . So in case i) u(x), x ∈ V has only finitely many values.
In case ii), we know by construction that u(x) is a norm one element of M .
Hence there are finitely many possibilities as well.
When C is an irreducible conic things are only slightly more involved.
The conic can be described by the equation
det(A, P, x) det(B, P, x)/ det(A, B, x)2 = δ
for some fixed δ ∈ K. Notice that for any x ∈ V we have αβ/γ 2 = δ. If
|αA|v 6= |βB|v fpr some v 6∈ S, we have necessarily
|γP|v ≥ max(|βB|v , |αA|v ).
Take squares and divide by |γP|2v = cv |βB|v |αA|v where cv = |P 2 /BA|v |δ|−1
v .
Hence
cv ≥ max(|βB/αA|v , |αA/βB|v )
We conclude that |β/α|v is 1 for almost all v 6∈ S. and that it has finitely
many possible values for the remaining v. Hence there are at most finitely
many possibilities for α/β, hence V is finite.

Remark. We expect that a similar theorem holds for higher degree rational
curves, but did not see how to prove this.

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1.3 The case ‘degree ≤ 3’
In this section we show that if deg(T ) ≤ 3, equation (T) has a Zariski dense
set of solutions under fairly mild conditions.
Theorem 1.3.1 Let L be a straight line in P2 defined over K. Then the
set of S-integral points on P2 \ L is Zariski dense in P2 .
Remark. The fact that for any straight line L there exists an S-integral
point on P2 \L is equivalent to saying that P2 \L is isomorphic over Spec(OS )
to the affine plane. The infinity of the number of S-integral points on P2 \ L
is then obvious.

Proof. Choose a point P ∈ L, defined over K. Let LP be any line through
P , distinct from L modulo every prime outside of S. According to the dual
version of Theorem 1.2.1 there exist infinitely many such lines. For any such
line LP there exist infinitely many S-integral points on LP \{P }. By varying
LP we find a Zariski dense set of S-integral points on P2 \ L. 2

Theorem 1.3.2 Let C be a geometrically irreducible conic defined over K.
By Kv we shall denote the completion of K with respect to a valuation v.
i) Suppose that C(Kv ) is empty for every place v ∈ S. Then there exist
at most finitely many S-integral points on P2 \ C.
ii) Suppose that for at least one place v ∈ S the equation C(Kv ) is not
empty. Then the set of S-rational points on P2 \ C is either empty or
Zariski dense in P2 .
Remark. Let us point out the necessity of the conditions in the case K = Q
and S = {∞}. The form X 2 + Y 2 + Z 2 does not represent zero in R = K∞
and sure enough, x2 + y 2 + z 2 = ±1 has only finitely many solutions. The
form 2X 2 + 5Y 2 − 5Z 2 does represent zero in R but 2x2 + 5y 2 − 5z 2 = ±1
does not have any solution, as one can see by looking modulo 5.
Remark. The ‘royal road’ to proving the above theorem would be to remark
that the condition on the places in S guarantees the existence of an infinite
projective automorphism group of P2 over Spec(OS ) which stabilises the
conic T (x, y, z) = 0. Images of our solution under the group elements yield
a Zariski dense set of solutions. However, the proof we present below uses a
shortcut, thus avoiding a discussion of non-ramifying primes from S in the
quaternion algebra associated to T .

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Proof. Part i). Let C be given by the equation T (x, y, z) = 0. For any
place v the function

|T (x, y, z)|v / max(|x|v , |y|v , |z|v )2

is continuous on P2 (Kv ) with the v-adic topology. Since P2 (Kv ) is compact,
our function has a minimum which we denote by mv . By assumption, we
have mv > 0 for any v ∈ S. Let now (x : y : z) be a solution of our ternary
form equation (T). For any v 6∈ S we have

|T (x, y, z)|v = Tv max(|x|v , |y|v , |z|v )2

where Tv is the v-adic valuation of the coefficient ideal of T . Note that
Tv = 1 with finitely many exceptions for v. When v ∈ S we have

where H(x, y, z) is the height of the point (x : y : z). We conclude from the
inequality that H(x, y, z) is bounded, hence the number of solutions to (T)
is finite.
Part ii) Denote the given S-integral point in P2 \ C by Q. Let Pv be
the v-adic point on C. For any v-adic neighbourhood U of Pv there exist
infinitely many lines through Q, defined over K and which intersect C in
U in a point whose coordinates are in a quadratic field. Let L be such a
line. It intersects C in two points, P, R say, whose coordinates are defined
over a quadratic extension M of K and such that P and R are conjugates.
By construction the place v splits in M . Hence, by Theorem 1.2.3, there
exist infinitely many S-integral points on L\{P, R}. Since we have infinitely
many choices for L we have proved our theorem. 2
The following theorem gives a positive answer to Question 6.10 asked
by J.Silverman in [Si] in 1984.

Theorem 1.3.3 Suppose C is a geometrically irreducible plane cubic curve.
Suppose C has at least one K-rational flex F . Suppose that the tangent at
C through F is not a component of C modulo any prime outside of S. Then
the set of S-integral points on P2 \ C is Zariski dense in P2 .

Remark. From examples it seems that the cardinality of C(K) has little
or no influence on the number of solutions. The curve given by 2x3 +
5y 3 + 7z 3 = 0 contains infinitely many rational points, but the equation
2x3 + 5y 3 + 7z 3 = ±1 has no solutions in Z, as can be seen by consideration
modulo 7. A similar example is given by the curve 2x3 +7y 3 +7z 3 = 0 which
even has the rational flex (0 : 1 : −1).

Proof. Let C be given by the equation T (x, y, z) = 0. Let L be the tangent
at C through F . Let M be a straight line through F , distinct from L modulo
every v 6∈ S. Let L(x, y, z), M (x, y, z) be linear forms corresponding to L
and M . Let R be an S-integral point on L \ F and let a = T (R)/M (R)3 .
Then the line L intersects the cubic T − aM 3 = 0 in the points F (triple)
and R. Hence L is a component of the cubic curve, i.e. T − aM 3 = L · Q
for some quadratic form Q. Notice that the conic Q = 0 intersects C in
only two points, namely the points of intersection of C and M distinct from
F . Let L, M, T be the ideals generated by the coefficients of L, M and T
respectively. The point R is S-integral on P2 \ M , hence (M (R)) = MR
where R is the ideal generated by the coordinates of R. So we obtain,
a ∈ T M−3 . Suppose there is a prime v 6∈ S such that |(a)T −1 M3 |v < 1.
Then we have modulo v that T ≡ L·Q, contradicting our assumption that L
is not a component of C modulo v. So we conclude that (a) = T M−3 . After
replacing of a−1 T by T we may as well assume that a = 1 and T = M3 .
Furthermore one checks that Q, the ideal generated by the coefficients of Q
satisfies
Q ⊂ L−1 T = M3 L−1 .
Consider the line Mλ given by M + λL = 0, where λ ∈ (3)−1 ML−1 . Then

14
P2 \ M we have that (M (P )) = MP, where P is the ideal generated by the
coordinates of P . Hence we can check that

λ ∈ (3)−1 QM−2 ⊂ (3)−1 ML−1 .

The conic Qλ = 0 intersects T = 0 in only two points, say Aλ , Bλ . We want
to show that P is distinct from Aλ , Bλ modulo any v 6∈ S. If |(λ)M−1 L|v ≤ 1
this is clear, the line Mλ is distinct from L modulo v and P stays away from
F , the intersection of L and Mλ . When v|3 and |(λ)M−1 L|v > 1 we note
that the conic Qλ reduces to L2 and the points Aλ , Bλ to F . Hence P , being
S-integral on L \ F , is again distinct from Aλ , Bλ .
We are now ready to apply Theorem 1.2.3 and prove our theorem. All
we have to do is point out that there exist infinitely many λ such that either
Aλ , Bλ are K-rational in the case |OS∗ | = ∞ or that Aλ , Bλ are conjugate
points defined over a quadratic extension of K in which at least one place
of S splits.
Let us make our choice of P more explicit. Let P0 be an S-integral
point on L \ F . Then any point of the form P = P0 + tF with t ∈ P0 F −1 is
S-integral on L \ F . Our choice for λ now becomes

λ = Q(P0 + tF )/3M (P0 )2 , t ∈ P0 F −1

hence λ is a quadratic polynomial in t. Furthermore, for any λ the discrim-
inant (up to a square in K) of the quadratic field over which Aλ , Bλ are
defined is given by a cubic polynomial D(λ) with leading coefficient Q(F ).
The zeros of D correspond precisely to the lines Mλ for which Aλ and Bλ
coincide. That is, the 2-torsion points when C is smooth and the double
point (twice) plus one extra point when C has a double point and finally, D
has a triple zero corresponding to the cusp when C has such a cusp. Now let
δ(t) = D(Q(P0 + tF )/3M (P0 )2 ). Notice that δ(t) is a polynomial of degree
6 whose leading coefficient, up to squares in K ∗ , is 3.
Suppose that 3−1 δ(t)√is the square of a polynomial in √ K[t]. Then
Aλ , Bλ are defined over K( 3). One easily checks that either 3 ∈√K and
hence |OS∗ | = ∞, or there is an infinite place which splits in K( 3)/K.
Since this holds for any t our theorem is proved in this case.
Now suppose that δ(t) has at least one zero of odd multiplicity. Then
there exist infinitely many t0 ∈ P0 F −1 such that δ(t0 ) is not a square in
K. If at least one infinite place of K is complex, we are done. So suppose
that all infinite places of K are real. Let v be such a place. By choosing t0
sufficiently large v-adically, we can see to it that δ(t0 ) is positive, hence v

15
p
splits in K( δ(t0 ))/K. Since there are infinitely many such possiblities for
t0 we are done. 2

Remark. The main idea in the above proof is the construction of excep-
tional conics in P2 \ C where C is the cubic given by T = 0. As remarked
before, there exist also infinitely many exceptional lines in P2 \ C. However,
I did not see how to use them in proving Zariski denseness of the solution
set of (T).
The only cases we have not dealt with yet are when C is reducible.
However, using the techniques from the previous section this is quite straight-
forward and we give only brief proofs, if any.

Theorem 1.3.4 Suppose C consists of two straight lines L1 , L2 which are
either defined over K or are conjugate lines over a quadratic extension M
of K. Let P be the point of intersection of L1 and L2 . Let V be the set of
S-integral points on P2 \ C.
Suppose that OS∗ is finite when the Li are defined over K and suppose
no place of S splits in M/K when the Li are not defined over K. Then V
is contained in a finite number of straight lines passing through P .
In all other cases V is either empty or Zariski dense in P2 .

i) If the Li pass through one point P , then V is contained in a finite set
of straight lines through P .

Now suppose that the Li do not pass through one point. Let A = L1 ∩L3 , B =
L2 ∩ L3 .

ii) Suppose either all Li defined over K or L1 , L2 conjugates over a quadratic
extension of K and L3 defined over K. If OS∗ is finite, the set V is
contained in a finite set of conics which pass through A and B, tangent
to L1 , L2 .

iii) Suppose L1 is defined over a cubic field and L2 , L3 are its conjugates,
or suppose that L1 , L2 are conjugates over a quadratic extension M of
K in which at least one place of S splits and OS∗ infinite, or suppose
that the Li are defined over K and OS∗ infinite. Then V is either empty
or Zariski dense in P2 .

16
Proof. Part i) follows simply from the fact that there are at most finitely
many S-integral points on a P1 minus three points.
Let P = L1 ∩ L2 . To prove part ii) we consider the function

det(A, P, x) det(B, P, x)
v(x) = .
det(A, B, x)2

Since any x ∈ V is S-integral on the lines Li we have that (v(x)) = P 2 /AB
for any x ∈ V . Furthermore, v(x) ∈ K. So, if OS∗ is finite, there are at most
finitely many possibilities for v(x) and V is contained in a finite set of conics
from the pencil det(A, P, x) det(B, P, x) = α det(A, B, x)2 .
To prove part iii) we construct automorphisms T as in Proposition 1.2.2.
In this case one easily verifies that the set of triples λ, µ, ν that can be cho-
sen lies Zariski dense in P2 . Hence, if x0 ∈ V , its transforms constitute a
Zariski dense set. 2

Theorem 1.3.6 Suppose C consists of a straight line L and an irreducible
conic Q, both defined over K. Let V be the set of S-integral points on P2 \C.

i) If OS∗ is finite, V is contained in a finite number of conics from the
pencil spanned by Q and 2L.

ii) Suppose that OS∗ is infinite and L intersects Q in (one or two) points
rational over K. Then V is either empty or Zariski dense in P2 .

iii) Suppose that OS∗ is infinite and L intersects Q in two conjugate points
over a quadratic extension M of K. If there exists P ∈ V such that
the tangents through P at Q are defined over K, the set V is Zariski
dense in P2 .

Proof. Consider the function u(x) = Q(x)/L(x)2 . For any x ∈ V we have
(Q(x)/L(x)2 ) = QL−2 . Hence, if OS∗ is finite, u(x), x ∈ V has only finitely
many values. This proves part i).
To prove part ii) assume that V is not empty and P ∈ V . Let A, B
be the points of intersection of Q and L. Assume that they are distinct.
As a consequence of Theorem 1.2.3 the line LAP through A and P contains
infinitely many points of V because LAP never reduces to a component of
Q or L. Let R be such a point and consider the line LBR through B and R.
On LBR we have again infinitely many points of V . The freedom of choice
of R gives us the Zariski denseness of V . Now assume that L is tangent to

17
Q with point of tangency A. Let Q0 be the conic from the pencil spanned by
Q and 2L which passes through P . Then there exist infinitely many points
on Q0 ∩ V . For each such point R we consider the line through A and R,
which again contains infinitely many points of V .
To prove part iii) let LP be a tangent of Q through P . Theorem 1.2.3
shows that V ∩ LP is infinite. Let R be such an S-integral point and let LR
be the tangent of Q through R, distinct from LP . Then LR is again defined
over K. Again we find that V ∩ LR is infinite. The freedom of choice of R
gives us the Zariski denseness of V . 2

Remark. As an application of part iii) of the previous theorem we note
that
(x2 + y 2 − z 2 )z = ±3k gcd(x, y, z) = 1, k ∈ Z≥0
has a Zariski dense set of solutions. We are in the case K = Q and S =
{∞, 3} and we can take P = (1 : 1 : 1).

1.4 The case ‘degree at least 4’
The only general case about which we can make a positive statement is when
T contains at least four distinct irreducible factors. The following theorem
is precisely Corollary 2.4.3 in Vojta’s book [V].
Theorem 1.4.1 (Vojta) Suppose T has at least four distinct irreducible
factors over K̄. Then the set of solutions to (T) is contained in a finite
union of plane algebraic curves.

Proof. Denote the four factors by T1 , T2 , T3 , T4 Without loss of generality
we may assume that these factors have coefficients in K. Let us also increase
S in such a way that OS has class number 1. Without loss of generality we
may then assume that the coefficients of each polynomial are integral and
generate OS . Equation (T) can be rewritten as

18
zero. We can assume that there is no subsum P = Q + R, where the
coefficients of Q and R have disjoint support, such that Q(T1 , T2 , T3 , T4 ) ≡
R(T1 , T2 , T3 , T4 ) ≡ 0. Extend S in such a way that all coefficients of P are
S-units.
Let
pi1 i2 i3 i4 X1i1 X2i2 X3i3 X4i4 .
X
P =
P
We define the weight of a term in P as j ij degTj . As a consequence of
our subsum condition we have that all terms of P have the same weight.
Consider the S-unit equation
X
pi1 i2 i3 i4 Ui1 i2 i3 i4 = 0 (4)

in the S-units Ui1 i2 i3 i4 . Because all terms of P have the same weight, each
equivalence class of solutions of (3) gives rise to an equivalence class of
solutions of (4) by putting

Ui1 i2 i3 i4 = T1 (x, y, z)i1 · · · T4 (x, y, z)i4 .

Conversely, given a solution Ui1 i2 i3 i4 we like to know which x, y, z correspond
to it in this way. It is well-known from results of Evertse and Van der
Poorten-Schlickewei [E2] that the solution set of (4) consists of a finite set
(of equivalence classes) and a possibly infinite set for which certain subsums
in (4) vanish. Each vanishing subsum in (4) defines a projective curve in
the x, y, z-plane. So, the infinite part of the solutions of (4) gives rise to a
finite union of curves in P2 on which (x, y, z) can lie. We now show that the
finitely many remaining solutions do the same. Let I, I 0 be two 4-tuples of
indices such that pI , pI 0 are not zero. A solution {UI }I of (4) gives rise to
an equation
i0 i0
UI T11 (x, y, z) · · · T44 (x, y, z) = UI 0 T1i1 (x, y, z) · · · T4i4 (x, y, z)

in x, y, z. Again this defines a nontrivial projective curve in P2 on which
(x, y, z) can lie. Since there are only finitely many such curves to be consid-
ered our proof is now concluded. 2

1.5 Exceptional curves
In this section we turn to the problem of the construction of exceptional
curves in P2 \ C. This turns out to be a hard geometrical problem and

19
we can only give some straightforward results. Let us forget all arithmetic
questions for the moment and take K = C. The set of exceptional curves
in P2 \ C will be called the exceptional set of C.
The only examples we have found of algebraic curves C of degree at
least 4 having an infinite exceptional set, arise as follows. Let αC1 + βC2
be a pencil of rational curves such that each element sits in the exceptional
set of any other element of the pencil. Take for C a union of components of
elements of this pencil. Examples of such pencils are

αX p Y q + βZ p+q = 0 α(Y Z p−1 + X p ) + βZ p = 0.

We also like to define very exceptional curves. They are images of
non-constant morphisms of C into P2 \ C. Similarly we can speak of the
very exceptional set of a curve C. For reducible curves we can describe the
very exceptional set.

Theorem 1.5.1 Let C be a plane algebraic curve without multiple compo-
nents. Suppose C consists of two components given by the projective equa-
tions F (x, y, z) = 0 and G(x, y, z) = 0 of degree f and g respectively. Let
m = f /(f, g) and n = g/(f, g). Then the generic element of the pencil
αF n + βGm = 0 is irreducible. If the very exceptional set of C is infinite,
every element in the pencil αF n + βGm = 0 is an exceptional curve or a
union of them (when the element is reducible).

Proof. The existence of a very exceptional curve E implies the existence of
three polynomials x, y, z ∈ C[t] such that F (x(t), y(t), z(t))G(x(t), y(t), z(t))
equals 1. Hence both F an G evaluated at this triple of polynomials are
constants and E is a component of αF n +βGm = 0 for suitable α, β. Because
of our choice of m and n the generic element of the pencil is irreducible. The
infinite cardinality of the exceptional set implies via Bertini’s theorem that
the generic element of the pencil is a rational curve. Since any element of
the pencil intersects any other element in at most one point, our theorem is
proved. 2
For non-rational irreducible curves of degree at least 4 we expect that
the (very) exceptional set is finite. A very preliminary result in this direction
might be the following theorem.

Theorem 1.5.2 Let C be an irreducible curve of genus at least 2. Then the
curves in the very exceptional set of C intersect C in at most finitely many
points.

20
Proof. Suppose that the very exceptional curve given by F (X, Y, Z) = 0
intersects C in a smooth point P . Let Q be another smooth point of in-
tersection with a very exceptional curve G = 0, say. Let dC , dF , dG be the
degrees of the curves involved. Notice that the function F dG /GdF is a ratio-
nal function on C with divisor dC dG dF (P − Q). Let C̃ be a normalisation
of C. Then P, Q and the function F dG /GdF lift to C̃. Let ψ : C̃ → Jac(C̃)
be the embedding of C̃ into its Jacobian via the map x ∈ C̃ 7→ x − P . Then
Q is mapped to Q − P and we have that dC dF dG (Q − P ) ∼ 0. By a theorem
of Raynaud [Ra] there exist only finitely many torsion points of Jac(C̃) on
the embedded curve C̃. Hence there are finitely many possibilities for Q. 2
For the next theorem we require a modified version of a result proved
independently by Voloch [Vol] and Brownawell-Masser [BM].

Let S be the set of distinct zeros of P1 · · · Pn , where we count ∞ as belonging
to S as soon as degPi 6= degPj for some i, j. Suppose that there exists no
proper subset I of {1, . . . , n} such that i∈I Pi (t) = 0. Then,
P

Suppose that the elements of I do not all lie on the same straight line and
suppose that (d/3, d/3, d/3) lies in the convex hull of I. If d > (3/2)|I|(|I| −
1) then there are only finitely many exceptional curves in P2 \ {T = 0}.

Proof. We must solve
T (x, y, z) = tm
in x, y, z ∈ C[t] and m ∈ Z≥0 . Without loss of generality we can assume that
(x, y, z) = 1. Suppose we have such a solution. Let n be the maximum of
the degrees of x, y, z. Let S be the set of distinct zeros of txyz plus the point
at ∞. Let ~e0 = (e1 (0), e2 (0), e3 (0)) be the numbers of factors t occurring in